list 选择列表的最后一个元素
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Choosing the last element of a list
提问by dondog
scala> last(List(1, 1, 2, 3, 5, 8))
res0: Int = 8
for having a result above, I wrote this code:
为了得到上面的结果，我写了这段代码：
val yum = args(0).toInt
val thrill:
def last(a: List[Int]): List[Int] = {
println(last(List(args(0).toInt).last)
}
What is the problem with this code?
这段代码有什么问题？
回答by michael.kebe
You can use last, which returns the last element or throws a NoSuchElementException, if the list is empty.
您可以使用last, 返回最后一个元素或抛出 a NoSuchElementException，如果列表为空。
scala> List(1, 2, 3).last
res0: Int = 3
If you do not know if the list is empty or not, you may consider using lastOption, which returns an Option.
如果您不知道列表是否为空，您可以考虑使用lastOption，它返回一个Option。
scala> List().lastOption
res1: Option[Nothing] = None
scala> List(1, 2, 3).lastOption
res2: Option[Int] = Some(3)
Your question is about List, but using laston a infinite collection (e.g. Stream.from(0)) can be dangerous and may result in an infinite loop.
您的问题是关于List，但last在无限集合（例如Stream.from(0)）上使用可能很危险，并可能导致无限循环。
回答by Jus12
Another version without using last(for whatever reason you might need it).
另一个不使用的版本last（无论出于何种原因您可能需要它）。
def last(L:List[Int]) = L(L.size-1)
回答by Pere Villega
You should better do:
你最好这样做：
val a = List(1,2,3) //your list
val last = a.reverse.head
Cleaner and less error-prone :)
更干净，更不容易出错:)
回答by Ravi
The recursive function last should following 2 properties. Your lastfunction doesn't have any of them.
递归函数最后应该遵循 2 个属性。你的最后一个函数没有它们中的任何一个。
Requirement #1. An exit condition that does not call recursive function further.
Requirement #2. A recursive call that reduces the elements that we began with.
要求#1。不进一步调用递归函数的退出条件。
要求#2。减少我们开始使用的元素的递归调用。
Here are the problems I see with other solutions.
以下是我在其他解决方案中看到的问题。
- Using built in function last might not be an option in interview questions.
- Reversing and head takes additional operations, which the interviewer might ask to reduce.
- What if this is a custom linked list without the size member?
- 使用内置函数 last 可能不是面试问题的选项。
- 反转和头部需要额外的操作，面试官可能会要求减少这些操作。
- 如果这是一个没有大小成员的自定义链表怎么办？
I will change it to as below.
我将其更改为如下所示。
def last(a: List[Int]): Int = a match {
//The below condition defines an end condition where further recursive calls will not be made. requirement #1
case x::Nil => x
//The below condition reduces the data - requirement#2 for a recursive function.
case x:: xs => last(xs)
}
last(List(1,2,3))
Result
结果
res0: Int = 3
回答by Marton Tatai
Albiet this is a very old question, it might come handy that the performance impact of head and last operations seems to be laid out here http://docs.scala-lang.org/overviews/collections/performance-characteristics.html.
尽管这是一个非常古老的问题，但 head 和 last 操作的性能影响似乎在这里列出http://docs.scala-lang.org/overviews/collections/performance-characteristics.html可能会派上用场。
回答by Sohum Sachdev
This is where the beauty of Scala comes to play!
这就是 Scala 之美发挥作用的地方！
val l: List[Int] = List(1,2,3,4)
val lastOption: Option[Int] = l.lastOption
By doing this, you will get an Optionof Int. Read more about Scala's Option's here.
通过这样做，你会得到一个Option的Int。在此处阅读有关 Scala 选项的更多信息。
Then finally, you can handle the Nonecase as you please:
最后，你可以随心所欲地处理这个None案例：
val last: Int = lastOption.getOrElse(0) //or however else you want to handle the case of an empty list
